Q. 28 - Calculus

Question

In Exercises 21–28 provide the first five terms of the series.

$\sum _{n=1}^{\infty }\frac{(-1{)}^n{n}^2}{n!}$

Step-by-Step Solution

Answer

Ans: The five terms of the series are $-1,\frac{4}{2!},-\frac{9}{3!},\frac{16}{4!},-\frac{25}{5!}$

Step 1. Given information:

$\sum _{n=1}^{\infty }\frac{(-1{)}^n{n}^2}{n!}$

Step 2. Finding the first term of the series:

The first term of the series $\sum _{n=1}^{\infty }\frac{(-1{)}^n{n}^2}{n!}$ is obtained by substituting $n=1$ in $\frac{(-1{)}^n{n}^2}{n!}$. Therefore, the value at $n=1$ is:

$\frac{(-1{)}^n{n}^2}{n!}=\frac{(-1{)}^1\left(1^2\right)}{1!}$ (Substituting)

$=\frac{-1}{1!}=-1$

Therefore, first term of the series $\sum _{n=1}^{\infty }\frac{(-1{)}^n{n}^2}{n!}$ is -1.

Step 3. Finding the second term of the series:

The second term of the series $\sum _{n=1}^{\infty }\frac{(-1{)}^n{n}^2}{n!}$ is obtained by substituting $n=2$ in $\frac{(-1{)}^n{n}^2}{n!}$. Therefore, the value at $n=2$ is:

$\frac{(-1{)}^n{n}^2}{n!}=\frac{(-1{)}^2(2{)}^2}{2!}$( Substituting)

$=\frac{4}{2!}$

The second term of the series $\sum _{n=1}^{\infty }\frac{(-1{)}^n{n}^2}{n!}$ is $\frac{4}{2!}$.

Step 4. Finding the third term of the series:

The third term of the series $\sum _{n=1}^{\infty }\frac{(-1{)}^n{n}^2}{n!}$ is obtained by substituting $n=3$ in $\frac{(-1{)}^n{n}^2}{n!}$. Therefore, the value at $n=3$ is:

$\frac{(-1{)}^n{n}^2}{n!}=\frac{(-1{)}^3(3{)}^2}{3!}$ (Substituting)

$=\frac{-9}{3!}$

The third term of the series $\sum _{n=1}^{\infty }\frac{(-1{)}^n{n}^2}{n!}$ is $\frac{-9}{3!}$.

Step 5. Finding the fourth term of the series:

The fourth term of the series $\sum _{n=1}^{\infty }\frac{(-1{)}^n{n}^2}{n!}$ is obtained by substituting $n=4$ in $\frac{(-1{)}^n{n}^2}{n!}$. Therefore, the value at $n=4$ is:

$\frac{(-1{)}^n{n}^2}{n!}=\frac{(-1{)}^4(4{)}^2}{4!}$( Substituting)

$=\frac{16}{4!}$

The fourth term of the series $\sum _{n=1}^{\infty }\frac{(-1{)}^n{n}^2}{n!}$ is$\frac{16}{4!}$.

Step 6. Finding the fifth term of the series:

The fifth term of the series $\sum _{n=1}^{\infty }\frac{(-1{)}^n{n}^2}{n!}$ is obtained by substituting $n=5$ in  $\frac{(-1{)}^n{n}^2}{n!}$. Therefore, the value at $n=5$ is:

$\frac{(-1{)}^n{n}^2}{n!}=\frac{(-1{)}^5(5{)}^2}{5!}$(Substituting)

$=-\frac{25}{5!}$

The fifth term of the series $\sum _{n=1}^{\infty }\frac{(-1{)}^n{n}^2}{n!}$ is $-\frac{25}{5!}$.